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Square root of a number using log

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For a given number find the square root using log function. Number may be int, float or double.

Examples: 

Input  : n = 9
Output : 3

Input  : n = 2.93
Output : 1.711724

We can find square root of a number using sqrt() method:

C++




// C++ program to demonstrate finding
// square root of a number using sqrt()
#include<bits/stdc++.h>
 
int main(void)
{
    double n = 12;
    printf("%lf ", sqrt(n));
    return 0;
}


Java




// Java program to demonstrate finding
// square root of a number using sqrt()
 
import java.io.*;
 
class GFG {
    public static void main (String[] args) {
    double n = 12;
    System.out.println(Math.sqrt(n));
 
 
// This code is contributed by akt_mit
    }
}


Python3




# Python3 program to demonstrate finding
# square root of a number using sqrt()
import math
 
if __name__=='__main__':
    n = 12
    print(math.sqrt(n))
 
# This code is contributed by
# Sanjit_Prasad


C#




// C# program to demonstrate finding
// square root of a number using sqrt()
using System;
 
class GFG
{
public static void Main()
{
    double n = 12;
    Console.Write(Math.Sqrt(n));
}
}
 
// This code is contributed
// by Akanksha Rai


PHP




<?php
// PHP program to demonstrate finding
// square root of a number using sqrt()
$n = 12;
echo sqrt($n);
 
// This code is contributed by jit_t
?>


Javascript




<script>
 
// Javascript program to demonstrate finding
// square root of a number using sqrt()
var n = 12;
 
document.write(Math.sqrt(n).toFixed(6));
 
// This code is contributed by aashish1995
 
</script>


Output

3.464102 

Time complexity: O(log2n), for using sqrt() function.
Auxiliary space: O(1)

We can also find square root using log2() library function: 

C++




// C++ program to demonstrate finding
// square root of a number using log2()
#include<bits/stdc++.h>
 
double squareRoot(double n)
{
    return pow(2, 0.5*log2(n));
}
 
int main(void)
{
    double n = 12;
    printf("%lf ", squareRoot(n));
    return 0;
}


Java




// Java program to demonstrate finding
// square root of a number using log2()
import java.io.*;
 
class GFG
{
static double squareRoot(double n)
{
    return Math.pow(2, 0.5 * (Math.log(n) /
                              Math.log(2)));
}
 
// Driver Code
public static void main (String[] args)
{
    double n = 12;
    System.out.println(squareRoot(n));
}
}
 
// This code is contributed by akt_mit


Python




# Python program to demonstrate finding
# square root of a number using sqrt()
import math
 
# function to return squareroot
def squareRoot(n):
 
    return pow(2, 0.5 * math.log2(n))
 
# Driver program
 
n = 12
print(squareRoot(n))
 
# This code is contributed by
# Sanjit_Prasad


C#




// C# program to demonstrate finding
// square root of a number using log2()
using System;
 
public class GFG{
     
static double squareRoot(double n)
{
     return Math.Pow(2, 0.5 * (Math.Log(n) /Math.Log(2)));
}
 
     
    static public void Main (){
            double n = 12;
            Console.WriteLine(squareRoot(n));
    }
//This code is contributed by akt_mit   
}


PHP




<?php
// PHP program to demonstrate finding
// square root of a number using log2()
function squareRoot($n)
{
    return pow(2, 0.5 * log($n, 2));
}
 
// Driver Code
$n = 12;
echo squareRoot($n);
     
// This code is contributed by ajit
?>


Javascript




<script>
    // Javascript program to demonstrate finding
    // square root of a number using log2()
     
    function squareRoot(n)
    {
         return Math.pow(2, 0.5 * (Math.log(n) /Math.log(2)));
    }
     
    let n = 12;
      document.write(squareRoot(n).toFixed(15));
 
</script>


Output

3.464102 

Time complexity: O(log2log2N), complexity of using log(N) is log(logN), and pow(x,N) is log(N), so pow(2,0.5*log(n)) will be log(logN).
Auxiliary space: O(1)

How does the above program work? 

 let d be our answer for input number n
 then n(1/2) = d 
     apply log2 on both sides
      log2(n(1/2)) = log2(d)
      log2(d) = 1/2 * log2(n)
      d = 2(1/2 * log2(n)) 
      d = pow(2, 0.5*log2(n))  

 



Last Updated : 25 Sep, 2022
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